A car comes to a bridge during a storm and finds the bridge washed out. The driv

Question

A car comes to a bridge during a storm and finds the bridge washed out. The driver must get to the other side, so he decides to try leaping it with his car. The side the car is on is 20.9 m above the river, while the opposite side is a mere 2.3 m above the river. The river itself is a raging torrent 69.0 m wide.

Part A How fast should the car be traveling just as it leaves the cliff in order just to clear the river and land safely on the opposite side?

Part B What is the speed of the car just before it lands safely on the other side?

Explanation / Answer

Ignoring air resistance and wind speed (and the fact that a car is not a point mass - moreover, it can be front wheel drive or rear wheel drive or all wheel drive);

- the car drops ( 20.9 - 2.3 = ) 18.6 meters
- with gravity (A) at 9.8 m/sec² that gives us a time of:
distance = A x t² : 2
t = ( 2 x 18.6 : 9.8 ) = 1.94 seconds
- so 69 m in 1.94 seconds or ( 69/1.94 x 36000/1000 = ) 128.04 km/h

( which is an underestimate, obviously; there is a dang STORM on)

The speed of the car just before it lands ...
- the horizontal speed is128.04 km/h (ignoring wind resistance and all that)
- the vertical speed is V = a t so 18.13 m/s or 65.27 km/h
- Treat it as a vector problem; Vcar = ( 128.04² + 65.27² ) = 143.71 km/h

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